Cambridge Additional Mathematics

Introduction to differential calculus (Chapter 13) 335

The concept of alimitis essential to differential calculus. We will see that calculating limits is necessary

for finding the gradient of a tangent to a curve at any point on the curve.

x 1 1 : 9 1 : 99 1 : 999 1 : 9999

f(x) 1 3 : 61 3 : 9601 3 :996 00 3 :999 60

The table alongside shows values for f(x)=x^2

wherexis less than 2 , but increasing and getting

closer and closer to 2.

We say that asxapproaches 2 from the left, f(x) approaches 4 from below.

x 3 2 : 1 2 : 01 2 : 001 2 : 0001

f(x) 9 4 : 41 4 : 0401 4 :004 00 4 :000 40

We can construct a similar table of values wherex

is greater than 2 , but decreasing and getting closer

and closer to 2 :

We say that asxapproaches 2 from the right, f(x) approaches 4 from above.

So, asxapproaches 2 from either direction, f(x) approaches a limit of 4. We write this as lim

x! 2

x^2 =4.

INFORMAL DEFINITION OF A LIMIT

The following definition of a limit is informal but adequate for the purposes of this course:

If f(x) can be made as close as we like to some real numberAby makingxsufficiently close to

(but not equal to)a, then we say that f(x) has alimitofAasxapproachesa, and we write

lim

x!a

f(x)=A.

In this case, f(x) is said toconvergetoAasxapproachesa.

Notice that the limit is defined forxclose to butnot equal toa. Whether the functionfis defined or not

at x=a is not important to the definition of the limit offasxapproachesa. Whatisimportant is the

behaviour of the function asxgetsvery close toa.

For example, if f(x)=^5 x+x

2

x

and we wish to find the limit asx! 0 , it is tempting for us to simply

substitutex=0into f(x). However, in doing this, not only do we get the meaningless value of^00 , but

also we destroy the basic limit method.

Observe that if f(x)=

5 x+x^2

x

=

x(5 +x)

x

then f(x)=

½

5+x if x 6 =0

is undefined if x=0.

The graph of y=f(x) is shown alongside. It is the straight

line y=x+5with the point (0,5) missing, called apoint of

discontinuityof the function.

However, even though this point is missing, thelimitof f(x) as

xapproaches 0 does exist. In particular, as x! 0 from either

direction, f(x)! 5.

We write lim

x! 0

5 x+x^2

x

=5 which reads:

“the limit asxapproaches 0 ,off(x)=

5 x+x^2

x

,is 5 ”.

A Limits

-5

5

missing

point

x

y

O

5x + x ___

f(x) = ^^^^^^

x___________________

2

4037 Cambridge

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Y:\HAESE\CAM4037\CamAdd_13\335CamAdd_13.cdr Friday, 4 April 2014 5:16:12 PM BRIAN